NABTEB GCE 2018/2019 CHEMISTRY ESSAY AND OBJ QUESTIONS AND ANSWERS – NOV/DEC EXPO

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2018 NABTEB GCE CHEMISTRY OBJ AND THEORY QUESTIONS AND ANSWERS SOLUTIONS – Nov/Dec Expo

 

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Expo

 

Exam Time: Friday 30th, Nov 2018
CHEMISTRY (Obj and Essay)
9:00am-11:30am

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NABTEB GCE CHEMISTRY obj and theory SOLUTIONS

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Chemistry OBJ:
1-10: DCDDACDCBA
11-20: CACDDADCBC
21-30: AABCDABCBA
31-40: CBACCAAAAB
41-50: CAAACAACDD

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(1ai)
(i) Alpha Radiations –> positively charged particles.
(ii) Beta Radiations –> Negatively charged particles.
(iii) Gamma Radiation –> Neutral(has no charge)

(1aii)
(i) Small amount of heat is involved in reaction involving chemical reaction; whereas greater amount of heat (energy) is liberated in nuclear reactions.
(ii) The nucleus of an atom is not affected in chemical reaction, whereas the nucleus of an atom is affected in nuclear reaction.

(1aiii)
Draw the chemical equation

(1b)
Given that X is 1s² 2S² 2P^6 3S² 3P^5
(i) Atomic number of X is 17
(ii) Group 7 or (VII)
(iii) They are electronegative elements. They ionize by gain of one electron.
(iv) X is chlorine

(1ci)
I. Because graphite contains carbon atoms arranged in layers with break forces between it’s layers, thereby making it slippery hence it is used as a lubricant.

II. Because diamond is the hardest substance : resulting from the giant crystalline forces in its molecule.

(1ciiii)
HCl(g) + NH3(g) –>NH4Cl

(1ciii)
Combination reaction/Neutralization.

(1civ)
A white solid at the upper part of the test tube.

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(2ai)
Arrhenius defined base as a substance which gives hydroxide ions, (OH-) in water.

(2aii)
NaOH

(2aiii)
(I) C12H22O11 – Non electrolyte.

(II) NH3 – Weak electrolyte.

(II) NaOH – Strong electrolyte.

(2aiv)
CH3COOH(aq) + KOH(aq) –> CH3COOK(aq) + H2O(l)

(2b)
HCl(aq) + NaOH(aq) –> NaCl(aq) + H2O(l)

Mole ratio 1 : 1
Ca = Molar conc of the hcl, 0.500moldm-³
Cb = Molar conc of NaOH, 0.3moldm-³
Vb = Volume of base, 20.00cm³
Va = ?
CaVa/CbVb = na/nb
Va = CbVbna/Canb
Va = 0.3×20×1/(0.5×1)
Va = 12cm³

(2c)
(i) NaOH – sodium hydroxide

(ii) Mg(HCO3)2 – Magnesium hydrogentrioxocarbonate(IV)

(2di)
Standard solution is one in with a concentration which is accurately known must be used to react with a solution of unknown concentration.

(2dii)
NaOH because its a soluble base.

(2e)
Fe(s) + 2HCl(g) —> FeCl2(s) + H2(g)

1 mole of Fe requires 2 mole of HCl for complete reaction.

Mole = molarity × volume in dm³
Mol = 2.2×0.05
Mol = 0.11mol of HCl
2mol of HCl requires 1 mol of Fe
0.11mol of HCl will require 0.55mol of Fe
Mol = mass/molar mass
Molar mass = mass/mole
Mass of Fe = 3.08g
Mole = 0.55mol
Molar mass of Fe = 3.08/0.55
Molar mass = 56gmol ^-1

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(3ai)
Soda lime is not deliquescent where caustic soda is deliquescent.

(3aii)
-It is partially soluble in water.
-It is a colourless and odourless gas.
-It is denser than air.

(3aiii)
(I) Emperical Formular:

Elements: C H
%composition: 82.76 17.24
Relative atomic mass: 12 1
Ratio: 82.76/12 17.24/1 6.8966 17.24

Ratio/smaller ratio = 6.8966/6.8966 17.24/6.8966

1 2.499
= 1 2.5

Then using a multiplication factor of 4, since the mole ratio did not turn out to become whole numbers;
1 mol of C × 4 = 4 mol of C
2.5 mol of H×4 = 10 mol of H
Hence, the empirical formula of the hydrocarbon is C4H10

(II) Finding the molecular formula of the hydrocarbon:
Recall that vapour density
V.D2 = Molecular mass, since V.D = 29, Hence MM = 292 = 58gmol^-1
Hence(C4H10)n = 58
(124+110)n = 58
(48+10)n = 58
58n = 58
n = 58/58 = 1
Hence mm = C4H10 –> molecular formula

(3bi)
Isomerism is the existence of two or more compounds with the same molecular formula but different structural formula.

(3bii)
Draw the chemical equation

(3biii)
Ethanol and dimethylether(methoxymetane)

(3biv)
Ethanol belongs to the alkanol family with higher boiling point WHILE dimethylether belongs to ether family ie different homologous series with lower boiling point.

(3c)
From the qualitative & quantitation analysis :
(i) The cation and anion present in a given compound.
(ii) The family of the elements of compound present in the given substance.
(iii) The amount or concentration of a given substance present in the solution or required to react with a known amount of a substance.

(3d)
Pass both Propene and Propane over bromine water, propene gas decolourizes bromine water while propane does not.

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(4a)
(i) Coal tar
(ii) Ammoniacal liquor (liquid)
(iii) Coal gas(gas)

(4b)
(i) Carbon(ii)oxide CO, carbon(iv)oxide CO2 and hydrogen gas H2(g)
(ii) This is because of the presence of carbon(II)oxide CO.

(4ci)
C(s) + O2(g) —> CO2(g)

(4cii)
(I) Physical Properties:
(i) It is a colourless gas.
(ii) It is an odourless gas.

(II) Chemical properties:
(i) It dissolves in water to form weak acid.
(ii) It reacts with alkali to form trioxocarbonate(iv)

(4di)
-Crystalline : Diamond and Graphite.
-Amorphous : Wood charcoal, Animal charcoal, Soot.

(4dii)
– Diamond is used in decoration and as Ostentation material.
– Graphite is used as a conductor/electrode.
– Wood charcoal is used in gas masking.
– Animal charcoal is used in decolorizing crude sugar.
– Soot is used in making polish.

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